To determine when Sara's train catches up to Rianna's train, we need to analyze the two equations provided for the distances traveled by both trains with respect to time.
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Rianna's Train Equation:
\( y = 40x + 10 \)
In this equation, \( y \) represents the distance traveled and \( x \) is the time in hours. The train leaves 15 minutes earlier, which is \( 0.25 \) hours; hence, after 15 minutes \( x = 0.25 \).To find the equivalent distance formula in terms of time for \( x = 0 \) (when Sara's train starts), we can substitute \( x = x + 0.25 \):
\( y = 40(x + 0.25) + 10 \)
Which simplifies to:
\( y = 40x + 10 + 10 = 40x + 20 \)
This accounts for the 15-minute head start that Rianna's train had. -
Sara's Train Equation:
\( y = 60x - 10 \)
Here, \( x \) is the time Sara's train has been traveling since it starts 10 minutes later, which is \( \frac{1}{6} \) hours.
Now we need to determine when the distances are equal to find when Sara's train catches up to Rianna's train.
Setting the equations equal to each other, we have:
\( 40x + 20 = 60x - 10 \)
Let's solve for \( x \):
\( 20 + 10 = 60x - 40x \)
\( 30 = 20x \)
\( x = \frac{30}{20} = 1.5 \) hours.
Now that we have the \( x \) value where they are equal, we need the \( y \) values to check which two lines represent this solution. We can substitute \( x = 1.5 \) back into either equation:
From Rianna's equation:
\( y = 40(1.5) + 20 = 60 + 20 = 80 \).
From Sara's equation:
\( y = 60(1.5) - 10 = 90 - 10 = 80 \).
If we had a graph with these equations plotted, the lines corresponding to these equations would intersect at the point (1.5, 80).
Now, looking at the options:
- The lines that correspond to the equations \( y = 40x + 20 \) and \( y = 60x - 10 \) will be the solution.
If we assume these options represent the following lines, we can refer back and determine:
- Given the equations, Line 1 is \( y = 40x + 20 \).
- Line 2 is \( y = 60x - 10 \).
Thus, the correct response is: Line 1 and Line 2.